Invert transform

at http://www.qntmpkt.blogspot.com

invert transform, an introduction

Refer to Aug. 24-th 2010 blog; "INVERT transform"

INVERT transform

The basic links at OEIS, http://www.tinyurl.com/4zq4q showing the
arxiv article listed - by Bernstein and Sloane.
The operations are presented in this blog, by way of example: find the INVERT
transform of (1, 2, 3,..).
...
1. Given an offset of 1 for S(x) = (x + 2x^2 + 3x^3 + 4x^4 +...); we plug about 6 of such terms into Wolfram alpha with S(x) in the following:
INVERT transform = (1 / (1 - S(x)) - 1.
So we plug into the field: 1/(1 - x - 2x^2 - 3x^3 - 4x^4 - 5x^5), enter.
Then subtract 1 getting (x + 3x^2 + 8x^3 + 21x^4 + 55x^5 + ...) where the coefficients = the even indexed Fibonacci numbers (1, 3, 8, 21,...). Therefore, the INVERT transform of (1, 2, 3,...) = (1, 3, 8, 21, 55,...).
....
The reverse operation = the INVERTi transform and can be defined as
(-1 / (1 + S(x)) + 1; and in an operational mode, we enter into WolframAlpha:
-1 / (1 + x + 3x^2 + 8x^3 + 21x^4 + 55x^5), then enter.
Subtract 1, getting (x + 2x^2 + 3x^3 + 4x^4 + ...). Therefore, the INVERTi
transform of (1, 3, 8, 21,...) = (1, 2, 3,...).
...
A property of INVERT transforms (say, B(x) is the INVERT transform of A(x);
then we preface B(x) with a 1 = (1 + B(x)) . Then A(x) * (1 + B(x)) = B(x).
That is, the INVERT transform shifts to the left when prefaced with a 1 and "convolved" with A(x). Example using A(x) = (1, 2, 3,...) and B(x) = (1, 3, 8,..).
...
Again, using WolframAlpha, we plug in:
(x + 2x^2 + 3x^3 + 4x^4 + 5x^5) * (1 + x + 3x^2 + 8x^3 + 21x^4 + 55x^5 + ...).
getting (x + 3x^2 + 8x^3 + 21x^4).
Thus, (1, 1, 3, 8, 21, 55,....) has shifted to the left to (1, 3, 8, 21,...) when
convolved with (1, 2, 3,..).

Proof of Adamson's conjecture on palindromic continued fractions, re: A179238

Proof of Adamson's conjecture: Gary W. Adamson, re: A179238 in OEIS stating that :

a. given t = irrational Tan A, then if t = the convergent to an infinitely periodic palindromic continued fraction, then Tan 2A is rational; and the converse.

b. Tan 2A is rational when Tan A is rational iff, Tan A = t = the convergent to an infinitely periodic palindromic continued fraction.

Example: barover[a], with a = 1 getting [1,1,1,1,...] = .618....; with the inverse 1.618.... = t (either case) = Tan A.; where 1.618... is irrational.. If t = .618...then Tan 2A = 2.0. If t = 1.618..., then Tan 2A = -2.
...
We will present the sections of the proof then put the separate components together.
Part A: Given t = Tan A, then the double angle 2A triangle hs lets of 2t and t^2-1, with hyptenuse t^2+1.
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The key step here is the relationship Q = t - 1/t = rational; example 1.618... - .618...= 1.
In order for t to be irrational but Tan 2A to be rational, t - 1/t = Q must be rational (to be shown); but for now continuing with this algrbra,; we have
Q = 1 - 1/t, then Qt = t^2 - 1, and
2t * (QT) = 2t * (t^2 - 1); with 2t / (t^2-1) = 2t / Qt = 2/Q; (and Q is rational).
...
Alternatively, using Tan 2A = 2TanA / (1 - Tan^2 A,) , let t = Tan A, say phi^(-3).
Then Tan 2A = 2*(phi^-3) / (1 - phi^-3) =
2 / (phi^3 - phi^(-3); fulfilling our required identity such that phi^3 - phi^(-3) = 4, since
phi^3 = 4.236...; where .236...= [4,4,4,4,4,4,....] ; palindromic
...Thus, a specific case. of Tan 2A = 2 Tan A / (1 - Tan^2 A), where Tan A = phi^3 or it's inverse.
...
Next, we investigate Part B:
States that t - 1/t must be rational, and looking at the "tail" of an infinitely periodic continued fraction (refer to Wolfram's Mathworld); the "tail" using the formula, and applying to barover [1,2,3] = [1,2,3,1,2,3,1,2,3,...] this is not palindromic and is thus a counter example. The "tail" becomes (2x + 7 / (3x + 10) setting this = to x, getting 3x^2 + 8x - 7, the roots to this being the convergent to barover [1,2,3].
...
We can straightaway obtain the formula 3x^2 + 8x - 7 by writing the partial quotients of [1,2,3] underneath as follows:
1,.......2,........3
1/1 2/3 7/10.
We put 2/3 and 7/10 into a 2x2 format as:
2, 7
3, 10 and perform the following operation such that lower left term (3) = a, upper right = c;
and (lower right - upper left) = b, or f(x), (a,b,c) = ax^2 + bx - c = 0 or
3x^2 + 8x - 7 = 0, and we note that a is not equal to c or 3 is not equal to 7.
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Next, it follows that in constants of the form used in our conjecture c - 1/c = rational; the corresponding quadratic equation must be such that a must = c.; this being the case, dividing through by (a, c) we obtain an equation of the form x^2 - bx - 1 = 0, where b is rational.
This equation is the characteristic polynomial for an infinitely periodic, palindromic continued fraction, equavalent to the statement in the above 2x2 format that the lower left term must equal the upper right term. In this case 3 is not equal to 7, and the roots are not of the form t - 1/t = rational.
....
Part C. In order for the lower left term to equal the upper right term, we again refer to the "tail" of an infinitely periodic continued fraction which in the case of barover[1,2,3] =
2x + 7 / (3x+10), then setting this to x, we get 3x^2 + 8x - 7 = 0; again 3 is not equal to 7.
We need the tail of the form where upper right and lower left terms are equal, in other words,
in the convergents to [1, 2, 3] we have the partial quotients underneath:
1,.......2,.......7 and
1,.......3,......10. since [1,2,3] = 7/10 and the partial quotient to the left = 2/3.
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Part D. We refer to Mathworld "Continued fraction", theorem 30 and 31 which state that:
pn / p(n-1) = [an, a(n-1), ....ao] = (30) and
qn / (q(n-1) = [an....a1]................=(31)
...In other words, (30) applies to the reversal of [a,b,c] saying we start with an, then procede to a0. For example, given [1,2,3,4] , theorem 30 uses [4, 3, 2, 1]. Where [1,2,3,4] = 30/43 as follows:
1,.......2,.......7,........30
1,.......3,......10,......43
...then reverse using [4, 3, 2, 1] = obtaining
1,.......3,.......7,.......10
4,......13,.....30,.....43.; so in essence, theorem 30 switches the position of the upper left and lower right terms, with the 7 and 43 terms invariant.
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Now if the reversal of [a,b,c] remains invariant and thus the upper right and lower left terms are unchanged, this could only occur if the reversal is palindromic, such as [1, 2, 2, 1] where our convergents are
1,....2,......5,......7
1,....3,......7,.....10.
Thus only in a palindromic continued fraction is the lower left term = to the upper right.
Part E.
It follows from the formula of the "tail" say given (using different a,b,c,...):
we set x =( ax + b) / (cx + d); but again changing the usage of our (a,b,c,...) the quadratic formula for the tail if the (upper right = lower left) condition holds must be such that in
ax^2 + bx - c, a must = c. where a,b,c are all rational.
...But only when in this quadratic, a = c does the result hold (dividing through by a = c), we obtain
x^2 - bx - 1 = 0 (with an allowance for changing signs); and if b is rational, then we have our necessary equation with roots t and 1/t such that for example 1.618... - .618 = 1, rational, or
t - 1/t is rational, but t is irrational.
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Combining all of the steps, Q.E.D. especially noting the key step "upper right = lower left term";
we have shown that in given t irrational, t must be the convergent of an infinitely periodic, palindromic continued fraction in order that angle 2A is rational. Of course, if Tan 2A is rational, t can be rational; but our conjecture only applies to the case in which Tan 2A is rational and t (Tan A) is irrational. The converse of the conjecture is also true.

Ternary Gray code

Enter "Gray code conversion rules" into Google to bring up the general rules; following rules in A147995 in OEIS: http://www.tinyurl.com/4zq4q
then, following the reflection principle for Gray code k, we write out (say k=3) 0,1,2; 2,0,1; 1,2,0; ...in the right column; i.e. we write each term in the series once, then for the next subset of 3, we repeat the last term recorded, or "reflect" it.
...0
...1
...2;
now reflect, starting the series again (2, then 0, 1,...):
...2,
...0,
...1;
now for the next subset we reflect the "1" getting
...1
...2 (continuing with 0,1,2, etc.)
Now for the next column going to the left, we mark down each term 3 times; and the next column after that (column 3), 9 times, so that each term in the series 0,1,2 (and generally, 0,1,2,3,...for any k) is marked down k^0, k^1, k^2,....times along with the reflection principle. Thus, for Ternary Gray code we obtain:
000
001
002
012
010
011
021
022
020
120
121
122
102
101
100
110
111
112
...

Gray code conversion rules

at http://www.tinyurl.com/4zq4q
then access sequence A147995
The reflection principle is as follows, by way of example, ternary Gray code :
First column, going right to left, within each set of 3 terms starting with 0, write 0,1,2, then the next set repeats the 2 (reflection) then continues with the series 0,1,2. Each set reflects, then continues in the series 0->1->2, so we obtain
0,1,2; 2,0,1; 1,2,0; 0,1,2;...; (but these are column 1 terms.). Column 2 terms each term is marked down 3 times: 0,0,0; then 1,1,1; 2,2,2; then reflect: 2,2,2;...
For column 3, each term is written down 9 times along with the reflection principle as before. Generally for Gray code K, each term is recorded K^0, K^1, K^2...times by columns. Putting the rules together for k=3, Ternary, we
obtain (for decimal 0, 1, 2, 3,....):
000
001
002
012
010
011
021
022
020
120
121
122
102
101
100
110
111
112
...
and, we note that the column change from n-th to (n+1)-th term =
A051064 in OEIS: (1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, ...) representing
the row (labeled 1,2,3,...) in the following multiplication table:
(heading terms not multiples of 3, * left column = powers of 3):
1,...2,...4,....5,....7,...8,...10...
3,..6..,12,..20, 21,.24,..30,..
9,..;
Then record the row that n=1,2,3,...occurs in, getting A051064 as before:
(1, 1, 2, 1, 1, 2, 1, 1, 3,...) = the ruler sequence for k=3.

Genomic matrices

Generalized Genomic Matrices, Silver Means, and Pythagorean Triples"> > is now published on the web;> http://www.scipress.org/journals/forma/frame/24.html>